Problem: What is the area of the region between the graphs of $f(x)=3e^x$ and $g(x)=-2$ from $x=0$ to $x=3$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3e^2}{2}+\dfrac{3}{e}-6$ (Choice B) B $3e^3 - 9$ (Choice C) C $\dfrac{3e^2}{2}+\dfrac{3}{e}+6$ (Choice D) D $3e^3 + 3$
Explanation: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${1}$ ${2}$ ${3}$ ${20}$ ${40}$ ${60}$ $f$ $g$ $y$ $x$ From the graph, it appears that $f(x)\ge g(x)$ between $x=0$ and $x=3$. From this we are looking to evaluate: $ \int_{0}^{3}\left( f(x)-g(x) \right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{0}^{3} \left( 3e^x- (-2) \right) \,dx \\\\ &= \int_{0}^{3} \left( 3e^x+2 \right) \,dx\\\\ &=3e^x + 2x~\Bigg|_{0}^{3} \\\\ &= \left( 3e^3+6 \right) -\left(3+0 \right)\\\\ &= 3e^3 + 3 \end{aligned}$ Answer The area is $3e^3 + 3$ square units.